Question

sin^-1(x) +sin^-1(1-x) =sin^-1√1-x²​

Answer 1

Answer:

x = {0, 1/2}

Step-by-step explanation:

Use the formula ;

${sin}^{ - 1} (x) + {sin}^{ - 1} (y) = {sin}^{ - 1} (x \sqrt{1 - {y}^{2} } + y \sqrt{ {1 - x}^{2} } )$

${sin}^{ - 1} (x) + {sin}^{ - 1} (1 - x) = \\ {sin}^{ - 1} (x \sqrt{1 - {(1 - x)}^{2} } + (1 - x) \sqrt{ {1 - x}^{2} } )$

${sin}^{ - 1} (x \sqrt{2x - {x}^{2} } + (1 - x) \sqrt{1 - {x}^{2} } ) = \\ {sin}^{ - 1} ( \sqrt{1 - {x}^{2} } )$

$x \sqrt{2x - {x}^{2} } + (1 - x) \sqrt{1 - {x}^{2} } = \\ \sqrt{1 - {x}^{2} }$

=

$x \sqrt{2x - {x}^{2} } = \sqrt{1 - {x}^{2} } (1 - (1 - x)$

=

$x \sqrt{2x - {x}^{2} } = x \sqrt{1 - {x}^{2} }$

Squaring, we get,

${x}^{2} (2x - {x}^{2} ) = {x}^{2}(1 - {x}^{2} ) \\ {x}^{2} (2x - {x}^{2} - 1 + {x}^{2} ) = 0 \\ => {x}^{2} (2x - 1) = 0 \\ => x = 0 \: \: or \: \: x = \frac{1}{2}$

Hence, x = 0 or x = 1/2.